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Mega Millions Method

Topic closed. 5 replies. Last post 5 years ago by jimmy4164.

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United States
Member #114352
July 29, 2011
1 Posts
Offline
Posted: July 29, 2011, 9:10 pm - IP Logged

I'm currently working on a method that I think can increase a person's chances of
winning the jackpot. Obviously it's impossible to guarentee a win unless you
buy every single possible combination. The method is extremely time consuming
and expensive and can only guarentee a minimum cash back amount. But once I
have it hashed out I was hoping I could post everything I have here and see if
anyone is willing to work with me on improving the method to further increase
the odds, cut down on the cost, and increase the minimum guarenteed
reimbursement.

Because of the current cost ($552 each time you play) the method is designed to be used only for lottery pools
and in games with large pay off amounts.

The idea is to target the 2 minimum prizes which are the Mega Only tickets and the
Mega plus 1 tickets.

The idea is to purchase 552 tickets in sets of 12. 12 tickets for each of the 46 mega #s. And within each set of 12,
52 numbers 1 through 56 played once and 4 numbers 1 through 56 played twice.
Within the 552 tickets 640 of the 1540 possible 2 digits combinations will be
repeated 3 times each and the other 900 repeated 4 times with no repeating 3, 4, or
5 digit combinations.

I'm currently making a chart of all winning lottery numbers from Sept 6, 1996 to
the present and comparing all the combinations. I'm only 1/2 through it but so
far it looks as if there has never been a repeat of any 5 + M combos, 5 number
combos, or 4 + M number combos. And so far I've only found a repeat of two 4
number combos and a small handful of 3 + M number combos.

I'm still working on the numbers but I believe the chances of winning the jackpot using this method
are 1 out of 325391.7333.

This
method will guarentee ONLY ONE of the following possible winning combinations:

One 5 + M, Four 1 + M, & Seven M only

One 5 + M, three 1 + M, & Eight M only

One 5 + M, two 1 + M, & Nine M only

One 5 + M, one 1 + M, & ten M only

One 5 + M & 11 M only

One 4 + M, one 2 + M, three 1 + M, & seven M only

One 4 + M, one 2 + M, two 1 + M, & eight M only

One 4 + M, one 2 + M, one 1 + M, & nine M only

One 4 + M, one 2 + M, & ten M only

One 4 + M, five 1 + M, & six M only

One 4 + M, four 1 + M, & seven M only

One 4 + M, three 1 + M, & eight M only

One 4 + M, two 1 + M, & nine M only

One 4 + M, one 1 + M & ten M only

Two 3 + M, three 1 + M, & seven M only

Two 3 + M, two 1 + M, & eight M only

Two 3 + M, one 1 + M, & nine M only

Two 3 + M & ten M only

One 3 + M, two 2 + M, two 1 + M, & seven M only

One 3 + M, two 2 + M, one 1 + M, & eight M only

One 3 + M, two 2 + M, & nine M only

One 3 + M, one 2 + M, one 1 + M, & nine M only

One 3 + M, one 2 + M, & ten M only

One 3 + M, two 1 + M, & nine M only

Four 2 + M, one 1 + M, & seven M only

Four 2 + M & eight M only

Three 2 + M, one 1 + M, and eight M only

Two 2 + M, two 1 + M, and eight M only

Two 2 + M, one 1 + M, and nine M only

One 2 + M, three 1 + M, and nine eight M only

Nine 1 + M & three M only

Eight 1 + M & four M only

Seven 1 + M & five M only

Six 1 + M & six M only

Or

Five 1 + M & seven M only

 

Please don't think I'm nuts.  I know this isn't a practical method in it's current state.  Once I have all the comparisons done and 552 test combinations picked out to use and compare against up coming winning combos I might have a better idea of how well it works and if there is a way to improve upon it.  If anyone else is interested in working on it, I'd be greatful for the help and am willing to share what I have so far with anyone else.

Thanks, MJ (TA)

    MADDOG10's avatar - smoke
    Beautiful Florida
    United States
    Member #5709
    July 18, 2004
    20124 Posts
    Offline
    Posted: July 29, 2011, 9:18 pm - IP Logged

    I'm currently working on a method that I think can increase a person's chances of
    winning the jackpot. Obviously it's impossible to guarentee a win unless you
    buy every single possible combination. The method is extremely time consuming
    and expensive and can only guarentee a minimum cash back amount. But once I
    have it hashed out I was hoping I could post everything I have here and see if
    anyone is willing to work with me on improving the method to further increase
    the odds, cut down on the cost, and increase the minimum guarenteed
    reimbursement.

    Because of the current cost ($552 each time you play) the method is designed to be used only for lottery pools
    and in games with large pay off amounts.

    The idea is to target the 2 minimum prizes which are the Mega Only tickets and the
    Mega plus 1 tickets.

    The idea is to purchase 552 tickets in sets of 12. 12 tickets for each of the 46 mega #s. And within each set of 12,
    52 numbers 1 through 56 played once and 4 numbers 1 through 56 played twice.
    Within the 552 tickets 640 of the 1540 possible 2 digits combinations will be
    repeated 3 times each and the other 900 repeated 4 times with no repeating 3, 4, or
    5 digit combinations.

    I'm currently making a chart of all winning lottery numbers from Sept 6, 1996 to
    the present and comparing all the combinations. I'm only 1/2 through it but so
    far it looks as if there has never been a repeat of any 5 + M combos, 5 number
    combos, or 4 + M number combos. And so far I've only found a repeat of two 4
    number combos and a small handful of 3 + M number combos.

    I'm still working on the numbers but I believe the chances of winning the jackpot using this method
    are 1 out of 325391.7333.

    This
    method will guarentee ONLY ONE of the following possible winning combinations:

    One 5 + M, Four 1 + M, & Seven M only

    One 5 + M, three 1 + M, & Eight M only

    One 5 + M, two 1 + M, & Nine M only

    One 5 + M, one 1 + M, & ten M only

    One 5 + M & 11 M only

    One 4 + M, one 2 + M, three 1 + M, & seven M only

    One 4 + M, one 2 + M, two 1 + M, & eight M only

    One 4 + M, one 2 + M, one 1 + M, & nine M only

    One 4 + M, one 2 + M, & ten M only

    One 4 + M, five 1 + M, & six M only

    One 4 + M, four 1 + M, & seven M only

    One 4 + M, three 1 + M, & eight M only

    One 4 + M, two 1 + M, & nine M only

    One 4 + M, one 1 + M & ten M only

    Two 3 + M, three 1 + M, & seven M only

    Two 3 + M, two 1 + M, & eight M only

    Two 3 + M, one 1 + M, & nine M only

    Two 3 + M & ten M only

    One 3 + M, two 2 + M, two 1 + M, & seven M only

    One 3 + M, two 2 + M, one 1 + M, & eight M only

    One 3 + M, two 2 + M, & nine M only

    One 3 + M, one 2 + M, one 1 + M, & nine M only

    One 3 + M, one 2 + M, & ten M only

    One 3 + M, two 1 + M, & nine M only

    Four 2 + M, one 1 + M, & seven M only

    Four 2 + M & eight M only

    Three 2 + M, one 1 + M, and eight M only

    Two 2 + M, two 1 + M, and eight M only

    Two 2 + M, one 1 + M, and nine M only

    One 2 + M, three 1 + M, and nine eight M only

    Nine 1 + M & three M only

    Eight 1 + M & four M only

    Seven 1 + M & five M only

    Six 1 + M & six M only

    Or

    Five 1 + M & seven M only

     

    Please don't think I'm nuts.  I know this isn't a practical method in it's current state.  Once I have all the comparisons done and 552 test combinations picked out to use and compare against up coming winning combos I might have a better idea of how well it works and if there is a way to improve upon it.  If anyone else is interested in working on it, I'd be greatful for the help and am willing to share what I have so far with anyone else.

    Thanks, MJ (TA)

    You're far from being nuts, as you say. Very practical if you ask me. I can see where you're targeting tickets in sets of 12. I think it's a good start. Keep up the good work and if I can help you in anyway, don't hesitate to ask.

    Welcome to LotteryPost..! 

                                                 

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      RJOh's avatar - chipmunk
      mid-Ohio
      United States
      Member #9
      March 24, 2001
      19831 Posts
      Offline
      Posted: July 31, 2011, 12:41 pm - IP Logged

      I'm currently working on a method that I think can increase a person's chances of
      winning the jackpot. Obviously it's impossible to guarentee a win unless you
      buy every single possible combination. The method is extremely time consuming
      and expensive and can only guarentee a minimum cash back amount. But once I
      have it hashed out I was hoping I could post everything I have here and see if
      anyone is willing to work with me on improving the method to further increase
      the odds, cut down on the cost, and increase the minimum guarenteed
      reimbursement.

      Because of the current cost ($552 each time you play) the method is designed to be used only for lottery pools
      and in games with large pay off amounts.

      The idea is to target the 2 minimum prizes which are the Mega Only tickets and the
      Mega plus 1 tickets.

      The idea is to purchase 552 tickets in sets of 12. 12 tickets for each of the 46 mega #s. And within each set of 12,
      52 numbers 1 through 56 played once and 4 numbers 1 through 56 played twice.
      Within the 552 tickets 640 of the 1540 possible 2 digits combinations will be
      repeated 3 times each and the other 900 repeated 4 times with no repeating 3, 4, or
      5 digit combinations.

      I'm currently making a chart of all winning lottery numbers from Sept 6, 1996 to
      the present and comparing all the combinations. I'm only 1/2 through it but so
      far it looks as if there has never been a repeat of any 5 + M combos, 5 number
      combos, or 4 + M number combos. And so far I've only found a repeat of two 4
      number combos and a small handful of 3 + M number combos.

      I'm still working on the numbers but I believe the chances of winning the jackpot using this method
      are 1 out of 325391.7333.

      This
      method will guarentee ONLY ONE of the following possible winning combinations:

      One 5 + M, Four 1 + M, & Seven M only

      One 5 + M, three 1 + M, & Eight M only

      One 5 + M, two 1 + M, & Nine M only

      One 5 + M, one 1 + M, & ten M only

      One 5 + M & 11 M only

      One 4 + M, one 2 + M, three 1 + M, & seven M only

      One 4 + M, one 2 + M, two 1 + M, & eight M only

      One 4 + M, one 2 + M, one 1 + M, & nine M only

      One 4 + M, one 2 + M, & ten M only

      One 4 + M, five 1 + M, & six M only

      One 4 + M, four 1 + M, & seven M only

      One 4 + M, three 1 + M, & eight M only

      One 4 + M, two 1 + M, & nine M only

      One 4 + M, one 1 + M & ten M only

      Two 3 + M, three 1 + M, & seven M only

      Two 3 + M, two 1 + M, & eight M only

      Two 3 + M, one 1 + M, & nine M only

      Two 3 + M & ten M only

      One 3 + M, two 2 + M, two 1 + M, & seven M only

      One 3 + M, two 2 + M, one 1 + M, & eight M only

      One 3 + M, two 2 + M, & nine M only

      One 3 + M, one 2 + M, one 1 + M, & nine M only

      One 3 + M, one 2 + M, & ten M only

      One 3 + M, two 1 + M, & nine M only

      Four 2 + M, one 1 + M, & seven M only

      Four 2 + M & eight M only

      Three 2 + M, one 1 + M, and eight M only

      Two 2 + M, two 1 + M, and eight M only

      Two 2 + M, one 1 + M, and nine M only

      One 2 + M, three 1 + M, and nine eight M only

      Nine 1 + M & three M only

      Eight 1 + M & four M only

      Seven 1 + M & five M only

      Six 1 + M & six M only

      Or

      Five 1 + M & seven M only

       

      Please don't think I'm nuts.  I know this isn't a practical method in it's current state.  Once I have all the comparisons done and 552 test combinations picked out to use and compare against up coming winning combos I might have a better idea of how well it works and if there is a way to improve upon it.  If anyone else is interested in working on it, I'd be greatful for the help and am willing to share what I have so far with anyone else.

      Thanks, MJ (TA)

      Because of the current cost ($552 each time you play) the method is designed to be used only for lottery pools
      The idea is to target the 2 minimum prizes which are the Mega Only tickets and the
      Mega plus 1 tickets.

      Since that are only 46 megaballs, targeting the minimum prizes of 0+1 or greater only cost $46 now, $552 allows you to repeat that strategy twelve times which guarantees a minimum of twelve 0+1 or greater prizes or possible more since the overall odds of winning a prize are 1:40.  You can guarantee a minimum match of 0+1 by playing all 46 megaballs but players are hoping to do better when they spend that much.

      I look forward to you posting your method once you have it hashed out, good luck to you.

       * you don't need to buy more tickets, just buy a winning ticket * 
         
                   Evil Looking       

        Avatar
        Kentucky
        United States
        Member #32652
        February 14, 2006
        7322 Posts
        Offline
        Posted: July 31, 2011, 11:36 pm - IP Logged

        Because of the current cost ($552 each time you play) the method is designed to be used only for lottery pools
        The idea is to target the 2 minimum prizes which are the Mega Only tickets and the
        Mega plus 1 tickets.

        Since that are only 46 megaballs, targeting the minimum prizes of 0+1 or greater only cost $46 now, $552 allows you to repeat that strategy twelve times which guarantees a minimum of twelve 0+1 or greater prizes or possible more since the overall odds of winning a prize are 1:40.  You can guarantee a minimum match of 0+1 by playing all 46 megaballs but players are hoping to do better when they spend that much.

        I look forward to you posting your method once you have it hashed out, good luck to you.

        I have used the 46 combo 2 if 5 wheel but never really experimented with running it 12 times because of the cost. It will be interesting so see which method MathJunkie uses.

          RJOh's avatar - chipmunk
          mid-Ohio
          United States
          Member #9
          March 24, 2001
          19831 Posts
          Offline
          Posted: August 1, 2011, 7:40 am - IP Logged

          I have used the 46 combo 2 if 5 wheel but never really experimented with running it 12 times because of the cost. It will be interesting so see which method MathJunkie uses.

          Once such strategies are consider seriously, it costs nothing to simulate them to see what would have happened.  It's just that most players don't simulate and study strategies they aren't thinking seriously of using. 

          It could be you might not use a 46 line wheel twelve times or twelve different ways but a 552 line wheel or some other strategy for picking the 552 lines of core numbers.

           * you don't need to buy more tickets, just buy a winning ticket * 
             
                       Evil Looking       


            United States
            Member #93947
            July 10, 2010
            2180 Posts
            Offline
            Posted: August 2, 2011, 12:54 am - IP Logged

            MathJunkie,

            The results of your strategy will probably be proportional to what is summarized here:

            http://www.lotterypost.com/thread/222395/1899924

            --Jimmy4164